\documentclass[11pt]{article} \hoffset=-0.05\textwidth \textwidth=1.1\textwidth \newcommand{\bndone}{87} \newcommand{\bndzero}{19} \bibliographystyle{amsalpha} \include{macros} \begin{document} N.B. Some sections here are written up formally; others are just notes. Note also I need to learn how to do bibliographies. \section{Introducing \Sha} For an elliptic curve $E$ defined over a number field $K$, let $\Sha(E/K)= \hbox{ker} (H^1(K, E) \rightarrow \prod_v H^1(K_v, E))$ be the Shafarevich-Tate group of $E/K$. Recall that $H^1(K, E)= \varinjlim H^1(L/K, E)$, where the direct limit is taken over all finite Galois extensions $L/K$; since each one of these groups is torsion, killed by $[L:K]$, so is $H^1(K, E)$. Therefore, $\Sha(E/K)$ is a torsion group, a fact we will later to use to conclude that $\Sha(E/K)=0$ for curves having $\Sha(E/K)[p]=0$ for all primes $p$. In what follows, we will obtain results of the form $\Sha(E/K)[p]=0$ for almost all $p$, where $[K:\Q]=2$. The canonical map $\Sha(E/\Q) \rightarrow \Sha(E/K)$ has kernel contained in $H^1(K/\Q, E)$, a finite abelian $2$-group, so $\Sha(E/K)[p]=0 \Rightarrow \Sha(E/\Q)[p]=0$ as long as $p \neq 2$. \section{Known Results for CM Elliptic Curves} NEED MORE SPECIFICALLY WHAT RUBIN GIVES-- The result of Rubin in \cite{rubin:main-conjectures} implies, for elliptic curves with CM by the ring of integers in a quadratic imaginary field $K$, the full BSD conjecture up to primes dividing the number of units in the ring of integers (at worst requiring $p$-descent checks for $p=2 \hbox{or} 3$). But does this hold for CM by any order, or just the ring of integers??? We will now summarize and give references for the main theoretical results that give information about $\Sha$ and the rank of elliptic curves over $\Q$. Progress was first made in the case of elliptic curves with complex multiplication: J. Coates and A. Wiles showed in \cite{coates-wiles:bsd} that for such an elliptic curve $E/\Q$, $L(E/\Q, 1)\neq 0 \Rightarrow E(\Q)$ is finite. Improving on their techniques, K. Rubin proved two major results in his paper \cite{rubin:cmsha} about elliptic curves over $\Q$ with complex multiplication. Rubin's Theorem $A$ implies that whenever $L(E/\Q, 1) \neq 0$, $\Sha(E/\Q)$ is finite. Theorem $A$ also includes a local result, providing evidence for the full Birch and Swinnerton-Dyer conjecture, describing for which primes $p$ $\Sha(E/\Q)[p]$ can be non-trivial. Note that before Rubin's result, not a single Shafarevich-Tate group was known to be finite; his local result makes $\Sha(E/\Q)$ effectively computable in some cases. Under the same hypotheses, Theorem $B$ of Rubin's paper states that $\hbox{rk}_\Z(E/\Q) \geq 2 \Rightarrow \hbox{ord}_{s=1}L(E/\Q, s)\geq 2$. The previous year B. Gross and D. Zagier proved (\cite{gross-zagier}) their limit formula \[L^{\prime}(E/K,1)= \frac{\iint_{E(\C)} \omega \wedge \overline{i \omega}}{\sqrt{D}} \hat{h}(y_K),\] where $y_K$ denotes the usual Heegner point. Combining Rubin's result with the work of Coates-Wiles and the theorem, of Gross-Zagier, one obtains the \begin{theorem} Let $E/\Q$ be an elliptic curve with complex multiplication by an order in a quadratic imaginary field. Then \[\hbox{ord}_{s=1}L(E/\Q, s) \leq 1 \Rightarrow \hbox{rk}_\Z(E/\Q)= \hbox{ord}_{s=1}L(E/\Q, s).\] \end{theorem} \section{Kolyvagin and Consequences} Kolyvagin shows (\cite{kolyvagin:euler_systems}) that if $y_K$ has infinite order (i.e., $L^{\prime}(E/K, 1)\neq 0$), then $E(K)$ has rank $1$. Moreover, he proves $\Sha(E/K)$ is finite, with the following bounds. Writing $C_D= [E(K): \Z y_K]$, and letting $c_3$ and $c_4$ be constants I don't understand yet that are mired in Kolyvagin's paper, $c_3C_D$ kills $\Sha(E/K)$, and $\# \Sha(E/K)| c_4C_D^2$. Under the same hypothesis, he obtains the corresponding result for $E/\Q$ and $E_D/\Q$: $2c_3C_D$ kills $\Sha(E/\Q)$ and $\Sha(E_D/\Q)$, and both groups have order dividing $c_42^aC_D^2$, where $a$ is the 2-rank of both groups. Moreover, he gives a condition for determining whether $E/\Q$ or its twist $E_D/\Q$ contains the point of infinite order. If $E$ has modular parametrization $\gamma : X_0(N) \rightarrow E$, then $\gamma \circ w_N = \varepsilon \gamma + \gamma(0)$, where $w_N$ is the principal (Fricke) involution on the modular curve (alternatively, $\varepsilon$ is the eigenvalue on the normalized newform associated to E of the involution $w_N$). Kolyvagin deduces that $\varepsilon=1 \Rightarrow$ $E(\Q)$ contains the point of infinite order, and $\varepsilon=-1 \Rightarrow$ $E_D(\Q)$ contains the point of infinite order. (Note that our calculations for $E/\Q$ will give us some info free of charge about $E_D/\Q$) $\varepsilon=1$ makes our calculations easier, because while we'll be using low-level $E/\Q$, we don't know how high the conductor of $E_D/\Q$ might be. In \cite{kolyvagin:weil} Kolyvagin shows that $23$ elliptic curves over $\Q$ have trivial Shafarevich-Tate group, and he verifies the full Birch and Swinnerton-Dyer conjecture for $5$ of them, reducing the verification in other cases to a computation we will check. Note, however, that all of these curves have rank $0$: they are quadratic twists of the rank $1$ curve $E: y^2= 4x^3-4x+1$ having $\varepsilon=1$. In particular, he shows for the $23$ curves $E_D: -Dy^2= 4x^3-4x+1$ with $D\in \{7,11,47,71,83,84,127,159,164,219,231,263,271,\\ 287,292,303,308,359,371,404,443,447,471\}$, $\Sha(E_D/\Q)=0$. These are the twists with $D<500$ (the extent of the tables of computations Kolyvagin had available), $D$ prime to the conductor of $E$ ($37$), and as usual ruling out $D=3,4$, for which $C_D=1$. This suffices to conclude that $\Sha(E_D/\Q)=0$ because in Theorem $B$ of his paper, Kolyvagin shows that for whichever of $E/\Q$ and $E_D/\Q$ that does not contain the point of infinite order, $\Sha$ is killed by $C_D$ (as opposed to $c_3C_D$, which works for both curves). By explicitly working with the curve $E$, Kolyvagin reduces the full BSD conjecture in this case to showing $r_2(D)=0$, where $r_2(D)= \#\{q\, \hbox{an odd prime}: q|D, (\frac{q}{37})=1, \hbox{and}\, a_q\, \hbox{is even}\}$. $(\frac{q}{37})$ denotes the Legendre symbol, and $a_q$ is the Hecke eigenvalue. NOTE Cremona's tables won't be good for handling these, because conductors may get very high. His outputs of $a_q$ also only go up to $100$, and some of our $D$'s have prime divisors $>100$. \section{Weakening the Hypotheses for Triviality of $\Sha(E/K)[p]$} S. Donnelly has explained a way to weaken the hypotheses of Gross's Proposition $2.1$, replacing $G(\Q(E[p])/\Q)= GL_2(/Z/\/Z)$ with the statement that $E$ admits no $p$-isogenies. Gross only uses the hypothesis on $G(\Q(E[p])/\Q)$ at two points in the paper: section $4$, where he constructs Kolyvagin's cohomology classes and shows that restriction gives an isomorphism $H^1(K, E[p]) \cong H^1(K_n, E[p])^{\mathcal{G}_n}$, where $K_n$ denotes the ring class field of conductor $n$ over $K$ and $\mathcal{G}_n= G(K_n/K)$; and section $9$, where he requires $H^n(K(E[p])/K, E[p])=0$ for all $n \geq 0$, for then by the Hochshild-Serre spectral sequence he obtains the isomorphism $H^1(K, E[p]) \cong H^1(K(E[p]), E[p])^{G(K(E[p])/K)}$. The following lemma (exercise $6$ in \cite{ribet-stein:serre}) will be useful in the next two propositions. \begin{lemma} There are maps \[G(\bar{\Q}/\Q) \rightarrow \hbox{Aut}(E[\ell]) \cong GL_2(\F_{\ell}) \rightarrow \F_{\ell}^*\] given by the representation $\rho_{\ell}$, a choice of basis, and the determinant. Their composition is surjective, giving the cyclotomic character $\chi_{\ell}: G(\bar{\Q}/\Q) \rightarrow \F_{\ell}^*$. \end{lemma} \proof The Weil pairing induces an isomorphism of $G(\bar{\Q}/\Q)$-modules $E[\ell] \bigwedge E[\ell] \cong \mu_{\ell}$. Let us fix a basis $\{e_1, e_2\}$ of $E[\ell]$, with respect to which $\rho_{\ell}(\sigma)$ has the form $\left( \begin{array}{cc} a & b \\ c & d \end{array} \right)$. Then $$\sigma(e_1 \wedge e_2)= (ae_1+ce_2) \wedge (be_1+de_2)= \hbox{det}(\rho_{\ell}(\sigma))e_1 \wedge e_2.$$ It follows that the above composition gives the cyclotomic character (i.e., the action of $G(\bar{\Q}/\Q)$ on $\mu_{\ell}$), which is clearly surjective. \endproof \begin{lemma} Let $p$ be an odd prime and $E$ an elliptic curve over $\Q$. If $E$ has no $\Q$-rational $p$-isogeny, then $E$ has no $K$-rational isogeny, where $[K: \Q]=2$. \end{lemma} \proof The existence of a $p$-isogeny over $\Q$ is equivalent to the reducibility of the representation \[\rho_{p}: G(\bar{\Q}/\Q) \rightarrow \hbox{Aut}(E[p]) \cong GL_2(\F_p).\] Let the image of this representation be $G$, and consider the corresponding representation \[\rho_{p}: G(\bar{\Q}/\K) \onto H \subset \hbox{Aut}(E[p]) \cong GL_2(\F_p).\] Then $[G:H] \leq 2$ (letting $\alpha \in G(\bar{\Q}/\Q)$ be a lift of complex conjugation in $G(K/\Q)$, $\rho_{p}(\alpha)$ must generate $G/H$, but clearly $\rho_{p}(\alpha)^2 \in H$). In particular, if $E[p]$ has a subgroup $P$ of order $p$ invariant under $H$, the action of $\alpha$ on $P$ can generate a subgroup of order either $p$ or $2p$. The latter is impossible, since $2p \nmid p^2$, and it follows that the action of all of $G$ leaves $P$ invariant. Thus, if the representation over $K$ is reducible, so is the representation over $\Q$, and we conclude that if $E$ has no $p$-isogenies over $\Q$, it cannot have any over $K$. \endproof The next proposition shows the weakened hypothesis is sufficient in section $9$ of Gross's paper. \begin{proposition} Suppose $E$ has no $\Q$-rational $p$-isogeny, and let $[K:\Q]=2$. Then $H^i(K(E[p])/K, E[p])=0$ for all $i >0$. \end{proposition} \proof First observe that $G(K(E[p])/K)=H$ is precisely the image of the Galois representation $\rho_p: G(\bar{\Q}/K) \rightarrow \hbox{Aut}(E[p]) \cong GL_2(\F_p).$ Let $G$ and $H$ be as in the preceding lemma. By the lemma, $H$ is the image of an irreducible representation. There are two cases to consider: if $p \nmid \#H$, the cohomology clearly vanishes because multiplication by $\#G$ kills the cohomology but is an isomorphism on the $p$-group $E[p]$. Otherwise, let $p|\#H$. By Proposition $15$ of \cite{serre:propgal}, $H$ either contains $SL_2(\F_p)$ or is contained in a Borel subgroup of $GL_2(\F_p)$. $E[p]$ is reducible under the action of a Borel subgroup, so by hypothesis $SL_2(\F_p) \subset H$. Thus, $H$ contains a nontrivial scalar (minus the identity); in fact, Lemma $4.1$ implies that $\hbox{det}: G \rightarrow \F_p^*$ is surjective, so $G= GL_2(\F_p)$ and contains all of the nontrivial scalars. As $H$ is a subgroup of index at most $2$, $H$ must contain at least half of the scalars (in particular, it must contain the squares in $\F_p^*$. Applying the inflation-restriction exact sequence to the subgroup of $G$ generated by the scalars implies that $H^i(G, E[p])=0$ for all $i>0$, because this subgroup has order prime to $p$ ($p-1$, in fact), and it leaves no subgroup of $E[p]$ invariant. \endproof The next proposition extends the result of section $4$ of Gross's paper. Note that if $E$ has no $p$-isogeny (defined over $\Q$) for $p>2$, then it has no $p$-torsion over $K$. \begin{proposition} Let $E$ be an elliptic curve with $E(K)[p]=0$, where $p>3$ or, if $p=3$, $K \neq \Q(\mu_3)$. Then $H^i(K_n/K, E(K_n)[p]) =0$ for all $i \geq 1$. \end{proposition} \proof We may write the abelian group $G(K_n/K)$ as a direct sum $P \oplus P^\prime$, where $P$ is its Sylow $p$-subgroup and $(p, \#{P^\prime})=1$. We claim that the subgroup of $E(K_n)[p]$ invariant under $P^\prime$ is trivial. Let $G = G(K_n/K)/H$, where $H$ is the subgroup of $G(K_n/K)$ that acts trivially on $E(K_n)[p]$. If $(\#G, p)=1$, $P\subseteq H$, so $P^\prime$ surjects onto $G$. As there is no nontrivial element of $E(K_n)[p]$ invariant under all of $G(K_n/K)$, the same then holds for $P^\prime$. If $p|\#G$, we cannot have $\#E(K_n)[p]= p$, for then a subgroup of $G$ of order $p$ would act trivially on $E(K_n)[p]$, contradicting the definition of $G$. Thus, $E(K_n)[p]$ is the full $p$-torsion subgroup of $E$, and we can identify $G$ with a subgroup of $GL_2(\Z/p\Z)$ acting on $E(K_n)[p]= (\Z/p\Z)^2$. We can choose a basis of $(\Z/p\Z)^2$ so that $G$ contains the subgroup $\left( \begin{array}{cc} 1 & x \\ 0 & 1 \end{array} \right)$, where $x \in \Z/p\Z$. Being abelian, $G$ must be contained in the normalizer of this subgroup, so $G \subseteq \{ \left( \begin{array}{cc} a & b \\ 0 & a \end{array} \right) | a,b \in \Z/p\Z \}$, and we claim that $G$ contains an element with $a \neq 1$. Since $E[p] \subset E(K_n)[p]$, the representation $G(\bar{\Q}/K) \rightarrow \hbox{Aut}(E[p])$ factors through $G(K_n/K)$ (recall that the image of the representation is $G(K(E[p]/K)$). We argued before that this image contained the scalars corresponding to squares in $\F_p^*$, so $P^\prime$ contains at least $\frac{p-1}{2}$ (which is $>1$ for $p>3$) elements that leave no subgroup of $E(K_n)[p]$ invariant. Now, the result will follow from an application of the inflation-restriction exact sequence: \[0\rightarrow H^1(P, E(K_n)[p]^{P^\prime}) \rightarrow H^1(K_n/K, E(K_n)[p]) \rightarrow H^1(P^{\prime}, E(K_n)[p])\] The first group vanishes since $E(K_n)[p]^{P^{\prime}}=0$, and the third group vanishes since the order of $P^{\prime}$ is prime to $p$, and thus to the order of $E(K_n)[p]$. We conclude that the middle group is trivial, as desired. We can therefore extend the sequence to the second cohomology groups, deduce the same triviality result, and by induction conclude that $H^m(K_n/K, E(K_n)[p])=0$ for all $m \geq 1$. If $p=3$, $E(K_n)[3]=E[3] \Rightarrow \mu_3 \subset K_n \Rightarrow K=\Q(\mu_3)$. The last implication holds because $G(K_n/\Q)$ is abelian since $G(K_n/K)$ and $G(K/\Q)$ are, so it has a unique index $2$ subgroup; both $K$ and $\Q(\mu_3)$ correspond to index $2$ subgroups by elementary Galois theory). This contradicts our assumption on $K$, so we must have $E(K_n)[3]=0$, in which case the cohomology is clearly trivial. \endproof \section{The Decomposition of $E(K)$ under Complex Conjugation} To compute the index $I_K= \#{E(K)/{\Z y_K}}$ needed to verify the full Birch and Swinnerton-Dyer conjecture, we compare the canonical height of the Heegner point $y_K$ with the height of a generator of $E(K)$, which we know to have algebraic rank 1 by Kolyvagin's theorem. To assist in finding the (infinite order) generator of $E(K)$, we use the \begin{proposition} For a quadratic imaginary field $K=\Q(\sqrt{D})$, $E(K)$ decomposes, up to $2$-torsion, as a direct sum $E(\Q) \oplus E_D(\Q)$, where $E_D$ is the quadratic twist of $E$. (Recall that from a Weierstrass equation of $E/\Q$, we obtain a Weierstrass equation of $E_D/\Q$ as follows: \[\begin{array}{c} E: y^2= x^3+ax+b \\ E_D: y^2= x^3+D^2ax+D^3b \end{array})\] \end{proposition} \proof Denote complex conjugation by $\tau$. We will decompose $E(K)$ into its eigenspaces under the action of $\tau$. First, we kill $E(K)[2]$ (by tensoring with $\Z[\frac{1}{2}]$, for instance), so that, using the fact that $\tau^2=1$, we can define the projections from $E(K)$ to its $+1$ and $-1$ eigenspaces under $\tau$: for $P\in E(K)$, \[P= \frac{1+\tau}{2}P + \frac{1-\tau}{2}P.\] But if $P=(x,y)\in E(K)$ satisfies $\tau P=P$, then $P\in E(\Q)$; if $\tau P=-P= (x, -y)$, then $x\in \Q$ and $y\in \sqrt{D}\Q$. In particular, $(Dx, D\sqrt{D}y)\in E_D(\Q)$, and conversely we can obtain any such point in the $-1$ eigenspace of $E(K)$ from a point of $E_D(\Q)$. Thus, having eliminated the $2$-torsion, the decomposition of $E(K)$ into its $\tau$-eigenspaces just reads \[E(K)= E(\Q) \oplus E_D(\Q).\] \endproof \section{Application of Gross-Zagier to our calculations} In light of Donnelly's observation, we can apply Kolyvagin's result that $\Sha(E/K)[p]=0$ for all odd primes $p$ not dividing the Heegner point $y_K$ (i.e., $p$ does not divide $C_D$) and such that $E/K$ has no $p$-isogenies. By the inflation-restriction sequence, $\Sha(E/\Q)[p]=$ for all of these primes as well, since $\Sha(E/\Q)[p]=$ maps to $\Sha(E/K)[p]$ with kernel contained in $H^1(K/\Q, E(K)$, a finite $2$-group. Thus, it will be essential to compute the constant $I_K= [E(K): \Z y_K]$ (actually, we only want this index for $E(K)/E(K)_{tors}$. First we will have to find a generator $P$ of the free part of $E(K)$ and compute its canonical height $\hat{h}(P)$. We can use another form of the Gross-Zagier formula to compute $\hat{h}(y_K)$: \[\hat{h}(y_K)= \frac{u^2 \sqrt{D}}{\|\omega_f\|} L^{\prime}(E_D/\Q, 1)L(E/\Q, 1),\] where $u$ is half the number of units in $K= \Q(\sqrt{-D})$, and $\|\omega_f\|$ is the Peterson norm of the newform $f$ corresponding to $E$ (note that for $D\neq 3, 4$, $u=1$). With these preliminaries, we can compute the index $I_K= \sqrt{\hat{h}(y_K)/{\hat{h}(P)}}$. Note that we are free to choose $K= \Q(\sqrt{-D})$ as long as $D\neq 3,4$ and all the prime factors of $N$ (the conductor of $E$) split in $K$. To find $D$ given $N$ amounts to solving a finite number of congruences, so we should be able to implement this. \bibliography{biblio} \end{document}